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quadratic equation class 10 extra questions

Solve the following quadratic equation for x: Let the required numbers be x and y, where x > y (d) 1, 3 ⇒ (7)2 = 4 × 7 × k y = 3 = \(\frac{x-2}{x+2}\) Let age of one of the two friends = x yr ∴ speed of train = 40 km/hour. Breadth of verandah = (2x + 12) m. Find the speed of stream. Question 5. (c) –\(\frac {4}{3}\) x = 2 ⇒ k ≥ 16 and 16 ≥ k. Given equation is: The roots of the equation 2x2 – 8x + c = 0 are eaual, the value of c is: Solution. speed to stream = 5 km/Hr. Find the speed of stream. ⇒ (2 + y)2 + y2 = 34 ⇒ x2 – 13x + 2x – 26 = 0 takes 1 hour more to go 32 km up stream than to return down stream to the same spot. (d) 2 If so, find its length and breadth. New speed of the train = (x + 5) km/Hr. (d) 8 Hence, base of the triangle = 12 cm km = \(\frac {360}{x}\) Hours (a) equal to 0, Question 14. as length is never negative (By fatorization method) ∴ Rehman’s age, 3 yr ago = (x – 3) yr ⇒ 2 × 30 × 30 = 9(225 – x2) If the equations have real roots, then D ≥ 0 Discriminant of (i) k2 – 256 and equation. i.e., speed of stream = 8 km/h. Given equation is = (-20)2 – 4 × 1 × 112 Product of roots = αß = (b – 2a)(b + 2a) Also, solve some extra questions at BYJU'S for better practice. ⇒ x(x – 12) + 5(x – 12) = 0 Answer: x + 2 x + 2 = x – 2 which is not true = (20 – x – 4) yr = (16 – x) yr β = b + 2a ∴ length of its height = (x + 7) cm x = -2, – \(\frac{1}{2}\) Question 15. ⇒ x2 + 44x – 33x – 1452 = 0 ⇒ (x – 12) (x + 5) = 0 (a) 1, -3, Question 3. as number of vessels are not negative 4x2 + 4bx – a2 + b2 = 0 (c) –\(\frac {7}{3}\) Here you will find Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Solution. (a) b > \(\frac {8}{5}\) The length and breadth of a room are 15 m and 12 m. There is a verandah surrounding the room and area of this is 90m2. (b) not changed = (7)2 – 4 × 9 × -2 ∴ x2 – (2b)x + (b2 – 4a2) = 0 According to the question, The difference of two numbers is 2 and the sum of their square is 34. y + \(\frac {3}{y}\) – 4 = 0 Is the following situation possible ? Amount spend = ₹ 2250 Let the shorter side QR of the rectangle = x m. 4 yr ago age of the other friend ∴ x = -4 Answer: Answer: y = 6 or y = -2 Question 8. If x2 – 5 = 0, then find the value of x: (b) not real ⇒ 360 × 5 = x2 + 5x as speed is never negative ⇒ x2 – 3x + 1 = 0 Let the breadth of the park = x metre ⇒ x(x + 300) – 250(x + 300) = 0 10x + y = 10(2) + 6 = 26. Question 14. ⇒ x (x + 11) = 132 × 11 (b) –\(\frac {3}{4}\) But this rectangle will be a square of side 20 m. Question 1. ∴ x = -a or -6. (d) 8, Question 16. ∴ x2 – (sum of roots) x + product of roots = 0 Solution. (b) p > – 2√5 = 2(Length + Breadth)=80 m Solution. ∴ x – 18 = 0 or x + 10 = 0 (a) a = 1 If aand Bare the roots of the equation ax2 + bx + b = 0, then prove that: Had the price of cloth been ₹ 50 per metre higher, then he would have been able to purchase the cloth less by 1.5 metre. Area of right angle ∆ = \(\frac {1}{2}\) × base × height Where ∴ b2 = 4ac [∴ Cost Price can never be negative] = Length + Breadth = 40 m ⇒ x(x +12) – 5(x + 12) = 0 ∴ x = 8 (d) none of these. Answer: (ii) 64 – 4k ⇒ x2 = 225 – 200 ⇒ 16x – x2 – 64 + 4x = 48 x = 3/2 Solution: (x + 1)(x – 2) + x = 0 ⇒ x 2 – x – 2 + x = 0 ⇒ x 2 – 2 = 0 D = b 2 – 4ac ⇒ (-4(1)(-2) = 8 > 0 ∴ Given equation has two distinct real roots. The required quadratic equation is ⇒ 30 = \(\frac {1}{2}\)x. x ≠ 0. ⇒ 2x2 = 800 and x2 – 8x + k = 0…(ii) ∴ x = 18 or x = – 10 (c) 2 ⇒ x = -120 or x = 20 (a) 1 = (36 – 35)2 = 1, Question 4. ⇒ 2x = -8 (∵ In rectangle every adjacent side makes an angle 90° to each other) If one root of x2 + kx + 3 = 0 is 1, then the value of k will be: Hence, k = 16. (d) a + b + c = 0, Question 9. ∴ roots are \(\frac{\sqrt{2}}{\sqrt{3}}\) or \(\frac{\sqrt{2}}{\sqrt{3}}\), Question 6. (a) real and unequal, Question 17. Discriminant of ax2 + bx + c = 0) is: So, it is possible to design the rectangular park of perimeter 80 m and area 400 m2. (a) 1 y = 1 = \(\frac{x-2}{x+2}\) ∴ roots are. Solution. Hence, the total number of birds is 576. Here a = 9, b = 7 and c= -2 (b) -3 ⇒ x(a + b + x) = – ab ∴ y = 3 or – 5 ⇒ y(y – 24) + 6(y – 24) = 0 (d) none of these. So, present age of Rehman = 7 yr. xy = 12 …..(i) (i) and (ii), we have ⇒ -7x + 126 √x + 1008 = 0 (a) one zero (b) ≥ 0 ∴ (-2) 2 + 2 (-2) – p = 0. (By factorization method) Solution. (b) b = 1 (b) -1, -2 ⇒ 3x2 + 16x + 16 = 4x2 + 12x + 8 Find the two numbers. Now, we first transform one of the radicals co the R.H.S. (a) 0 and length of cloth \(\frac{2250}{250}\) = 9 metre. or x2 – 7x + 10 = 0 (a) 6 But (∵ breadth never be negative) Then, age of other friend = (20 – x) yr Find the number. as speed is never negative (a), Question 6. On comparing the above equation with ax2 + bx + c = 0, we get 2016. A quadratic equation ax2 + bx + c = 0, a ≠ 0 has equal roots if b2 – 4ac is: ∴ length of base = 5 cm. Find the value of k. ⇒ 64x = 576 – x2 a = 1, b = -20 and c = 112 ∴ x = 6 Now, new price of cloth per metre become = ₹ (x + 50) Hence, the solutions are -2, –\(\frac{1}{2}\), \(\frac{3 \pm \sqrt{5}}{2}\). \(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6. Question 12. Solution. (d) none of these. [∵ (a + b)2 = a2 + 2ab + b2] Solve: 2x4 – x3 – 11x2 + x + 2 = 0. Nature of root, D = b2 – 4ac = (4)2 – 4 × 1 × 0 = 16 > 0 ⇒ (2x + b + a) (2x + b – a) = 0, Question 5. If so, determine their present ages. ∴ Length of cloth purchase (b) ∴ Put y = 6 in (i) we get ⇒ y2 = 8x …..(ii) (a) equal to 0 x2 – (90x – 30x) – 2700 = 0 You can also download Class 10 Maths to help you to revise complete syllabus and score more marks in your examinations. 3x + 10 = 36 + 6 – x – 12\(\sqrt{6-x}\) Answer: 9x2 + 49y2 – 42xy and x2 + y2 = 34 …(ii) Determine whether the quadratic equation 9x2 + 7x – 2 = 0 has real roots. (d) -5. Let = \(\frac{x-2}{x+2}\) = y (c) -2√5 < p < 2√5 Question 3. Since, x ≠ -44 because speed can’t be negative. ⇒ y2 = 8 × 18 = 144 ⇒ x2 – 4x – 8 = 0 Find the measure of the cloth purchased and price per metre. Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations. (b) not changed, Question 18. The product of the roots of quadratic equation 3x2 – 4x = 0) is: Given – 5 is a root of quadratic equation 2x2 + px – 15 = 0 ⇒ (x – 18)(x + 10) = 0 ⇒ y2 – 24y + 6y – 144 = 0 ⇒ x(x – 90) + 30(x – 90) = 0 The two digit number is Answer: (b) -3, Question 23. ⇒ x2 + (12 – 5)x – 60 = 0 The required equation is is, Question 8. If the roots of 5x2 – px + 1 = 0 are real and distinct, then: So, breadth of the rectangle = 90 m and length of the rectangle = 90 + 30 = 120 m. Question 7. ⇒ (2x)2 + 2.2xb + (b)2-(a)2 = 0 According to the question, Solution. Let Numbers be x and y, such x > y (b) b < \(\frac {-8}{5}\) Question 1. (c) –\(\frac {4}{3}\), Question 21. Time taken by passenger train to cover Now substitute x + \(\frac {1}{x}\) = y and speed of boat in up stream = (24 – x) km/h (d) a + b + c = 0 ⇒ (x – 7)(x + 3) = 0 ⇒ 2x2 – 4x – x + 2 = 0 (c) ≤ 0 (b) +2 ⇒ x(x – 18) + 10(x – 18) = 0 Solution. Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers, Question 1. Solve the equation ∴ x = -300 or x = 250 ⇒ y2 + 2y – 15 = 0 2x = 3 Divide both sides by x2, ∴ speed of boat in down stream = (15 + x) km/H According to the question, Answer: ⇒  x = 20 If the discriminant (D) of a quadratic equation ax2 + bx + c = 0, a ≠ 0 is greater than zero, the roots are:

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